This problem tests understanding of accumulation function behavior when the integrand equals zero without changing sign. Since $F'(x) = f(x)$ by the Fundamental Theorem, critical points occur where $f(x) = 0$. At $x = 2$, we have $f(2) = 0$ with no sign change, meaning $F$ has a critical point but no local extremum. The function continues its previous trend (increasing since $f(x) > 0$ before $x = 2$) without creating a maximum or minimum. Choice B would require a sign change from positive to negative. When the integrand touches zero without changing sign, the accumulation function has neither a local maximum nor minimum.

This problem tests understanding of accumulation function monotonicity when the integrand is always positive. Since $G'(x) = g(x) > 0$ for all $x$, the accumulation function $G$ has a positive derivative everywhere, making it increasing throughout its domain. The fact that $g(3) = g(5)$ affects the concavity but not the monotonicity of $G$. Even when the integrand has equal values at different points, as long as it remains positive, the accumulation function continues increasing. Choice B incorrectly suggests constant behavior. As long as the integrand remains positive, accumulation functions are always increasing.

This problem requires understanding how extrema in the integrand affect inflection points in accumulation functions. Since $G'(x) = q(x)$ and $G''(x) = q'(x)$ by differentiation, inflection points in $G$ occur where $q'(x) = 0$ with a sign change. At $x = 5$, $q$ has a local maximum, which means $q'(5) = 0$ and $q'(x)$ changes from positive to negative. This creates a sign change in $G''(x)$, producing an inflection point in $G$ at $x = 5$. Choice A incorrectly suggests a local maximum, but extrema in $G$ require $q(x) = 0$, not extrema in $q$. When the integrand has extrema, the accumulation function has inflection points.

This problem requires understanding how the integrand's monotonicity affects the accumulation function's concavity. Since $A'(x) = g(x)$ and $A''(x) = g'(x)$ by differentiation, the concavity of $A$ depends on whether $g'(x)$ is positive or negative. Given that $g$ is decreasing on $(0,5)$, we have $g'(x) < 0$ throughout this interval. This means $A''(x) < 0$, so $A$ is concave down on $(0,5)$. Choice B incorrectly suggests concave up, which would require $g$ to be increasing rather than decreasing. When analyzing accumulation function concavity, examine whether the integrand is increasing or decreasing.

This problem tests our ability to locate extrema in accumulation functions with varying lower limits. Since $F'(x) = r(x)$ by the Fundamental Theorem, critical points occur where $r(x) = 0$. Given that $r(x) > 0$ for $x < 1$ and $r(x) < 0$ for $x > 1$, the integrand changes from positive to negative at $x = 1$. This means $F'(x)$ changes from positive to negative, indicating $F$ has a local maximum at $x = 1$. Choice A incorrectly identifies $x = 3$, but $x = 3$ is the lower limit where $F(3) = 0$ by definition. For accumulation functions, extrema occur where the integrand changes sign, regardless of the lower limit value.

This problem involves analyzing accumulation functions when both the integrand and its derivative are negative. Since $A'(x) = f(x) < 0$ on $(0,7)$, the function $A$ is decreasing throughout this interval. For concavity, $A''(x) = f'(x) < 0$, so $A$ is concave down on $(0,7)$. Both conditions being negative creates an accumulation function that decreases at an increasing rate. Choice B would be incorrect since concave up requires $f'(x) > 0$. When both the integrand and its derivative are negative, the accumulation function decreases with decreasing rate of change (accelerating downward).

This problem requires understanding how monotonicity changes in the integrand create inflection points in accumulation functions. Since $G'(x) = p(x)$ and $G''(x) = p'(x)$ by differentiation, inflection points occur where $p'(x) = 0$ with a sign change. Given that $p$ increases on $(-3,1)$ and decreases on $(1,5)$, we have $p'(x) > 0$ before $x = 1$ and $p'(x) < 0$ after $x = 1$. This means $G''(x)$ changes sign at $x = 1$, creating an inflection point. Choice D incorrectly suggests all zeros of $p$ create inflection points, but inflection points require sign changes in $p'$. Inflection points in accumulation functions occur where the integrand changes from increasing to decreasing or vice versa.

This problem involves analyzing accumulation functions with constant integrands. Since $A'(x) = f(x) = 3$ for all $x$ in $[-2,2]$, the accumulation function has a constant positive derivative. This means $A$ is increasing at a constant rate of 3 units per unit of $x$, making $A$ linear with slope 3. Choice B would be incorrect since the constant is positive, not negative. When the integrand is a positive constant, the accumulation function increases linearly at that constant rate.

This problem tests understanding of how an integrand's monotonicity affects the accumulation function's concavity. Since F'(x) = f(x) and F''(x) = f'(x), the fact that f is decreasing on (1,5) means f'(x) < 0, which makes F''(x) < 0. Therefore, F is concave down on (1,5). Students often confuse the integrand being decreasing with the accumulation function being decreasing, but a decreasing integrand actually determines the concavity, not the monotonicity, of F. The key principle is that a decreasing integrand produces a concave down accumulation function.

This question examines critical points of accumulation functions and their relationship to sign changes in the integrand. Since H'(x) = h(x), when h(x) = 0 at x = 3, we have H'(3) = 0, making x = 3 a critical point of H. Because h changes from negative to positive at x = 3, H' changes from negative to positive there, indicating H transitions from decreasing to increasing. This means H has a local minimum at x = 3. Students might confuse this with a maximum by reversing the sign analysis. Remember that when the integrand changes from negative to positive, the accumulation function reaches a local minimum.